Question: $(x_n)_{n=1}^\infty$ is a sequence with $x_n\neq0 $ for all $n$, also let $x_n$ tend to infinity. Let $(y_n)_{n=1}^\infty$ be defined by $y_n=\frac{1}{x_n}$, show that this converges to zero.

Definition for tending to infinity:$\forall K \in \mathbb{R} \exists N\in \mathbb{N} \forall n \in \mathbb{N} ,n>N:x_n>K $

Definition for convergence to zero:$\forall \varepsilon > 0 \exists N\in \mathbb{N} \forall n \in \mathbb{N} ,n>N:∣x_n∣<\varepsilon $

Idea: I know that $x_n$ tends to infinity since this is assumed. Then it must satisfy the defintion, therefore there must exist an $N$ which satisfies the defintion which i will call $N_x$ to prevent confusion.

To prove that $y_n$ converges to zero then I should choose $N_y=\lceil\frac{1}{N_x}\rceil$

Proof (attempt): Given $\varepsilon>0$ choose $N_y=\lceil\frac{1}{N_x}\rceil$. Given $n \in \mathbb{N}$, $n>N$ we have $∣y_n∣$$=∣\frac{1}{x_n}∣\leq \frac{1}{N_x}\leq\varepsilon $

Not entirely sure my selection for $N_y$ is correct, I understand why the statement is true however the proving put I am struggling.

  • $\begingroup$ For any $\varepsilon >0$ there exists $M \in \Bbb{N}$ such that $1/M< \varepsilon$ solves this, as $1/n$ is obviously decreasing $\endgroup$ – B.Swan May 11 at 20:53

Take $\varepsilon>0$. Since $\lim_{n\to\infty}x_n=\infty$, there is some $N\in\mathbb N$ such that $n\geqslant N\implies x_n>\frac1\varepsilon$. But then $n\geqslant N\implies 0<\frac1{x_n}<\varepsilon$. In particular, $\left\lvert\frac1{x_n}\right\rvert<\varepsilon$.

  • $\begingroup$ Hi thank you for your reply, would you mind clarifying how it implies that $x_n>\frac{1}{\varepsilon}$? $\endgroup$ – ViB May 11 at 21:17
  • $\begingroup$ I am using the definition of $\lim_{n\to\infty}x_n=\infty$, taking $K=\frac1\varepsilon$. Is it clear now? $\endgroup$ – José Carlos Santos May 11 at 21:21

Your intuition is right: the definition of $N_y$ is a bit off.

When you choose $N_x$, this value is for a particular $K$, from the "For all $K$ there exists $N\in\mathbb{N}$" part of the first statement. As $K$ increases without bound, so will $N_x$. Here, I think you're choosing $N_y$ with the goal of showing that $y_n$ is sufficiently small for all $n\geq N_y$. But if you let $K\rightarrow\infty$, choose a corresponding $N_x$, and then let $N_y=\lceil\tfrac{1}{N_x}\rceil$, this will result in $N_x$ shooting off to $\infty$ but $N_y$ will end up being $1$ all the time. Your goal will then be proving that $y_n$ is sufficiently small for all $n\geq N_y=1$, which won't be true.

For this problem, you've been given a relationship between the values of $\{x_n\}$ and $\{y_n\}$, so you should use this relationship to somehow connect the formal statements of "$x_n\rightarrow\infty$" and "$y_n\rightarrow 0$". Let's restate your proof goal:

Choose any $\epsilon>0$ (this choice is out of your control, your proof must work for any choice). You want to produce an $N\in\mathbb{N}$ such that $\{y_n\}$ satisfies the definition.

First, pick some $K>0$ (you're allowed to do this since we know that $x_n\rightarrow 0$) such that if $x_n>K$ then $y_n<\epsilon$. This $K$ will be expressed algebraically in terms of $\epsilon$. (EDIT: I think you responded to another post very recently with this choice of $K$ - use that one.) Then, if you take some $N_x$ such that $x_n>K$ for all $n\geq N_x$, then you'll be able to say something about $N_x$ and $\{y_n\}$ similarly.


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